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S5 Physics waves paper 3: refraction between media and uses of total internal reflection
S5 Physics waves paper 3: refraction between media and uses of total internal reflection
Ten items cover Snell law between two media, the inverse link between index and speed, optical fibres and normal incidence.
Direct answers Items 01 to 02 cover refraction across media and optical fibres, 03 to 05 critical angle judgements and normal incidence, 06 to 09 wavelength conversion, index ratios and the glass to water critical angle, and 10 the confusion of sine with tangent.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S5 Physics Waves and Refraction |
| Level | S5 |
02 Questions (10 items)
- FoundationLight passes from glass of index 1.5 into water of index 1.33 at 30 degrees. Find the angle of refraction.
- IntermediateOptical fibres carry light along a curved path. Which principle makes this work?
- IntermediateA medium has a critical angle of 45 degrees. Find its refractive index.
- IntermediateLight of 45 degrees leaves a medium of index 1.6 towards air. What happens?
- ChallengeLight strikes a boundary at normal incidence. What is the angle of refraction?
- FoundationA wave of 500 Hz travels at 8 m/s in deep water and 6 m/s in shallow water. Find both wavelengths in cm.
- ChallengeLight has three quarters of the vacuum speed in medium A and nine sixteenths in medium B. Find the ratio of their refractive indices.
- ChallengeGlass has index 1.5 and water 1.33. Find the critical angle for light going from glass into water.
- FoundationA wave has wavelength 0.25 m and frequency 1 360 Hz. Find its speed.
- Error redoFor index 1.5 and 30 degrees incidence a student uses tangent and gets 21.1 degrees. Find the correct angle.
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | about 34.3 degrees | sin theta2 is 1.5 times sin 30 degrees divided by 1.33, giving 0.5639. |
| 2 | Total internal reflection | At the core boundary the angle exceeds the critical angle, so the light is fully reflected back into the core. |
| 3 | about 1.414 | n is 1 divided by sin 45 degrees, that is 1 divided by 0.7071. |
| 4 | Total internal reflection, since the critical angle is about 38.7 degrees | Work out the critical angle first and compare it with the angle of incidence. |
| 5 | Zero, the ray is not deviated | sin zero is zero, so the refracted angle is zero as well and only speed and wavelength change. |
| 6 | 1.6 cm and 1.2 cm | Wavelength is speed divided by frequency, 8 over 500 and 6 over 500, then converted to cm. |
| 7 | 3 to 4 | Index is inversely proportional to speed, so a speed ratio of 4 to 3 gives an index ratio of 3 to 4. |
| 8 | about 62.5 degrees | sin C is 1.33 divided by 1.5, which is 0.8867. |
| 9 | 340 m/s | 0.25 times 1 360 is 340. |
| 10 | about 19.5 degrees | Snell law uses sine, not tangent, so sin theta2 is sin 30 degrees divided by 1.5, giving 0.3333. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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